Derivative of Field-Operator and Vector-Potential

[Abigale]

Hello,

I regard field-operators, whereby $\Psi$ is a fermionic-annihilation-operator and

$\vec{A(\vec{r})}$ is an electromagnetic-vector-potential.

Is it possible to do the following step?


$$
\nabla \vec{A} \Psi =\Psi \nabla \vec{A} + \vec{A}\nabla\Psi
$$

And if its correct, why?

Thx
Abby

 

[Abigale]

Ok I think I got a Solution.
If I consider Coulomb Gauge

$$
\nabla \vec{A}=0
$$
and I can write
$$
\nabla \vec{A} \Psi = \vec{A}\nabla\Psi
$$

 

[Avodyne]

$$
\nabla\cdot( \vec{A} \Psi) =(\nabla\cdot \vec{A})\Psi + \vec{A}\cdot\nabla\Psi
$$
by the product rule for derivatives. If you like, you can move $\Psi$ to the left in the first term, since $\Psi$ and $\vec A$ commute.